Two corners of an isosceles triangle are at $(8 ,5 )$ and $(6 ,1 )$ . If the triangle's area is $15$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-06-07T11:18:32

The length of three sides of triangle are $2sqrt5 ,5sqrt2 ,5sqrt2$ unit

Base of the isocelles triangle is $B= sqrt((x_2-x_1)^2+(y_2-y_1)^2)) = sqrt((8-6)^2+(5-1)^2)) =sqrt(4+16)=sqrt20 =2sqrt5$ unit

We know area of triangle is $A_t =1/2*B*H$ Where $H$ is altitude.
$:. 15=1/cancel2*cancel2sqrt5*H or H= 15/sqrt5$ unit

Legs are $L = sqrt(H^2+(B/2)^2)= sqrt(( 15/sqrt5)^2+((cancel2sqrt5)/cancel2)^2)=sqrt(45+5) = sqrt 50 = 5sqrt2$ unit

The length of three sides of triangle are $2sqrt5 ,5sqrt2 ,5sqrt2$ unit [Ans]

Two corners of an isosceles triangle are at (8 ,5 )(8 ,5 ) and (6 ,1 )(6 ,1 ). If the triangle's area is 15 15 , what are the lengths of the triangle's sides?